Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Question
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Chapter 28, Problem 53P

(a)

To determine

The ground state energy of the electron.

(a)

Expert Solution
Check Mark

Answer to Problem 53P

The ground state energy of the electron is 0.067 eV.

Explanation of Solution

The energy of the photon emitted during transition from the first excited state to the ground state is 0.20 eV.

Write the expression for energy of the quantized level in terms of the ground state energy.

En=n2E1                                                            (I)

Here, En is the energy of the level n, n is the quantum number and E1 is the energy of the ground state.

Substituting 2 for n in (II)

E2=22E1                                                           (II)

Here, E2 is the energy of the first excited state.

Write the expression for energy difference between the first excited state and the ground state.

ΔE=E2E1                                                          (III)

Here, ΔE is the energy difference.

Substituting (II) in (III)

ΔE=22E1E1=3E1                                                          (IV)

Rearranging (IV)

E1=ΔE3                                                           (V)

Substituting 0.20 eV for ΔE in (V) to find E1

E1=0.20 eV3=0.067 eV

Thus, the ground state energy is 0.067 eV.

(b)

To determine

The possible energies of the emitted photon if the electron is initially in the third excited state.

(b)

Expert Solution
Check Mark

Answer to Problem 53P

The possible energies of the emitted photon in the increasing order are 0.20 eV, 0.33 eV, 0.47 eV, 0.53 eV, 0.80 eV and 1.0 eV.

Explanation of Solution

If the initial state of the electron is 4, The possible transitions are 43, 32, 21, 42, 41, and 31. The energy differences between these levels are the possible energies of the emitted photon.

Substitute 4 and 3 in (I) to find E4 and E3  respectively.

E4=42E1=16E1                                                           (VI)

E3=32E1=9E1                                                           (VII)

Subtracting the corresponding energies in the allowed transition to find the energy of the photon and substitute 0.067 eV for E1

For 43, ΔE=16E19E1=7E1=7×0.067 eV=0.47 eV.

For 32, ΔE=9E14E1=5E1=5×0.067 eV=0.33 eV.

For 21, ΔE=4E1E1=3E1=3×0.067 eV=0.20 eV.

For 42, ΔE=16E14E1=12E1=12×0.067 eV=0.80 eV.

For 41, ΔE=16E1E1=15E1=15×0.067 eV=1.0 eV.

For 31, ΔE=9E1E1=8E1=8×0.067 eV=0.53 eV.

Thus, the possible energies of the emitted photon in the increasing order are 0.20 eV, 0.33 eV, 0.47 eV, 0.53 eV, 0.80 eV and 1.0 eV.

(c)

To determine

The plot of the wave function of the electron in the third excited state.

(c)

Expert Solution
Check Mark

Answer to Problem 53P

The wave function is a sign wave with three nodes.

Explanation of Solution

Write the expression for the wave function of a particle in a box.

ψn=2Lsin(nπxL)                                                    (VIII)

Here, ψn is the wave function of the quantum state n, L is the length of the box  and x is the position co-ordinate.

Substitute 4 for n in (VIII) to find ψ4

ψ4=2Lsin(4πxL)                                                     (VIII)

The plot of the wave function versus position is given below:

Physics, Chapter 28, Problem 53P

(d)

To determine

The effect of increasing the length of the box on its energy.

(d)

Expert Solution
Check Mark

Answer to Problem 53P

The energy spacing between two levels decreases.

Explanation of Solution

Write the expression for the energy of a particle in a box.

En=n2h28mL2                                                    (VIII)

Here, m is the mass of the electron and h is the Planck’s constant.

En is inversely proportional to the square of the length of the box. Thus, increasing the length decreases the energy of each level and hence the energy spacing also decreases.

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Chapter 28 Solutions

Physics

Ch. 28 - Prob. 2CQCh. 28 - Prob. 3CQCh. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQCh. 28 - Prob. 8CQCh. 28 - Prob. 9CQCh. 28 - Prob. 10CQCh. 28 - Prob. 11CQCh. 28 - Prob. 12CQCh. 28 - Prob. 13CQCh. 28 - Prob. 14CQCh. 28 - Prob. 15CQCh. 28 - Prob. 16CQCh. 28 - Prob. 17CQCh. 28 - Prob. 18CQCh. 28 - Prob. 1MCQCh. 28 - Prob. 2MCQCh. 28 - Prob. 3MCQCh. 28 - Prob. 4MCQCh. 28 - Prob. 5MCQCh. 28 - Prob. 6MCQCh. 28 - Prob. 7MCQCh. 28 - Prob. 8MCQCh. 28 - Prob. 9MCQCh. 28 - Prob. 10MCQCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 5PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Prob. 13PCh. 28 - Prob. 15PCh. 28 - Prob. 14PCh. 28 - Prob. 17PCh. 28 - Prob. 16PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 23PCh. 28 - Prob. 22PCh. 28 - Prob. 25PCh. 28 - Prob. 24PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 28PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 32PCh. 28 - Prob. 31PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 39PCh. 28 - Prob. 41PCh. 28 - Prob. 40PCh. 28 - Prob. 38PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - Prob. 45PCh. 28 - Prob. 46PCh. 28 - Prob. 47PCh. 28 - Prob. 48PCh. 28 - Prob. 49PCh. 28 - Prob. 50PCh. 28 - Prob. 51PCh. 28 - Prob. 52PCh. 28 - Prob. 53PCh. 28 - Prob. 54PCh. 28 - Prob. 55PCh. 28 - Prob. 56PCh. 28 - Prob. 57PCh. 28 - Prob. 58PCh. 28 - Prob. 59PCh. 28 - Prob. 60PCh. 28 - Prob. 61PCh. 28 - Prob. 62PCh. 28 - Prob. 63PCh. 28 - Prob. 65PCh. 28 - Prob. 64PCh. 28 - Prob. 66PCh. 28 - Prob. 67PCh. 28 - Prob. 68PCh. 28 - Prob. 69PCh. 28 - Prob. 70PCh. 28 - Prob. 71PCh. 28 - Prob. 72PCh. 28 - Prob. 73PCh. 28 - Prob. 74PCh. 28 - Prob. 75PCh. 28 - Prob. 76PCh. 28 - Prob. 77PCh. 28 - Prob. 79PCh. 28 - Prob. 78PCh. 28 - Prob. 80PCh. 28 - Prob. 81PCh. 28 - Prob. 82PCh. 28 - Prob. 83PCh. 28 - Prob. 84P
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